933 9.5 Nonlinear Systems of Equations The two equations form this system. x2y = 75 (1) x2 + 4xy = 85 (2) Step 4 Solve the system. We solve equation (1) for y to obtain y = 75 x2 . x2 + 4xy = 85 (2) x2 + 4x a 75 x2 b = 85 Let y = 75 x2 . x2 + 300 x = 85 Multiply. x3 + 300 = 85x Multiply by x, where x ≠0. x3 - 85x + 300 = 0 Subtract 85x. We are restricted to positive values for x, and considering the nature of the problem, any solution should be relatively small. By the rational zeros theorem, factors of 300 are the only possible rational solutions. Using synthetic division, as shown in the margin, we see that 5 is a solution. Therefore, one value of x is 5, and y = 75 52 = 3. We must now solve x2 + 5x - 60 = 0 for any other possible positive solutions. Use the quadratic formula. x = -5 + 252 - 41121-602 2112 ≈5.639 Quadratic formula with a = 1, b = 5, c = -60 This value of x leads to y ≈2.359. Step 5 State the answer. There are two possible answers. First answer: length = width = 5 in.; height = 3 in. Second answer: length = width ≈5.639 in.; height ≈2.359 in. Step 6 Check. See Exercise 69. S Now Try Exercises 67 and 69. 5)1 0 -85 300 5 25 -300 1 5 -60 0 (++)++* Coefficients of a quadratic polynomial factor 9.5 Exercises CONCEPT PREVIEW Answer each of the following. When appropriate, fill in the blank to correctly complete the sentence. 1. The following nonlinear system has two solutions, one of which is 13, 2. x + y = 7 x2 + y2 = 25 2. The following nonlinear system has two solutions with real components, one of which is 12, 2. y = x2 + 6 x2 - y2 = -96 3. The following nonlinear system has two solutions, one of which is 1 , 32. 2x + y = 1 x2 + y2 = 10 Find the positive solution.
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